# Recent posts in Preparation Advice

1
You should know the precedence rule. $\to$ has higher precedence than quantification.
2
For empty string as language, q0 as final states gives 16 DFAs. But when both q0 and q1 are final states we must ensure none of the transitions go to q1 from q0. This reduces the possible DFAs to $4$ as we have 2 choices for each {0, 1} transitions from q1 (either to q0 or to q1 itself). So, totally 16 + 4 = 20 DFAs.
3
according to my understanding, there are only 4 pages in the main memory (which should be 0,1,2,3) and then each page is 16 bytes (so, 16*4=64 bytes) should be the total main memory size. then how come there are pages 4 and 5 and how could address generated be more than 64 (6 bits). there is seriously something wrong with my concept. please can anyone help.?
4
$\lim_{x\rightarrow \frac{\pi}{2}}{( \sin x)}^{\tan x}$
5
Cache memory size must include the size of tag memory also, though here it is almost negligible.
6
Thanku sir..
7
In this question why there is 1 cache miss for the first 64 bytes of array ..Plz explain.....
8
Yeah why not you have found a solution it cab be one of the option but always choose the best one
9
hi I have already posted the same thing . Please have a look ! My question is different !
10

I am Nitish Gupta and I secured AIR-13 with a score of 950/1000 in GATE CS 2020. I am in the final year of my Graduation at CCET, Chandigarh and therefore this was my first attempt. I’d like to share the key points related to my preparation so that it can help future aspirants.

I joined a classroom course at a GATE Coaching GATEGURUS here in Chandigarh and most of the theory was covered by the classroom coaching. But doesn’t matter if you join a well-known coaching, online coaching or study using standard books, once you collect the theory from these sources, it all comes...

11
'exactly one' is a subset of 'atleast one'. Both are not equivalent.
12

Those who want to prepare for coding rounds for various colleges/company interviews, please check out this link.

13
Link Utilization $=\dfrac{\text{Amount of data sent}}{\text{Max. amount of data that could be sent}}$ Let $x$ ... $L=40\times 64\ bits=40\times \dfrac{64}{8}\ bytes=40\times 8\ bytes=320\text{ bytes (answer)}$
14
(A) P-Home, Q-Power, R-Defense, S-Telecom, T-Finance : X (R wants either Home or Finance or no portfolio) (B) R-Home, S-Power, P-Defense, Q-Telecom, T-Finance: Valid (c) P-Home, Q-Power, T-Defense, S-Telecom, U-Finance: X (U does not want if S gets one of the five) (d) Q-Home, U-Power, T-Defense, R-Telecom, P-Finance: X( R wants either Home or Finance or no portfolio)
15
The total number of nodes accessed including root will be 5. The order is, (9)-->(5)-->(5,7)-->(9,11)-->(13,15).
16
Which one of the following is the recurrence equation for the worst case time complexity of the quick sort algorithm for sorting $n$ ( $\geq$ 2) numbers? In the recurrence equations given in the options below, $c$ is a constant. $T(n) = 2 T (n/2) + cn$ $T(n) = T ( n - 1) + T(1) + cn$ $T(n) = 2T ( n - 1) + cn$ $T(n) = T (n/2) + cn$
17
3! * S(4, 3) = 3! * 6 = 36 https://gateoverflow.in/5596/no-of-surjective-functions http://www.math.uiuc.edu/~kostochk/math413/comb.pdf
18
Correct Option: D $foo$ is printing the lowest digit. But the $printf$ inside it is after the recursive call. This forces the output to be in reverse order $2, 0, 4, 8$ The final value $sum$ printed will be $0$ as $C$ uses pass by value and hence the modified value inside $foo$ won't be visible inside $main$.
19
Given 32 bits representation. So, the maximum precision can be 32 bits (In 32-bit IEEE representation, maximum precision is 24 bits but we take best case here). This means approximately 10 digits. A = 2.0 * 1030, C = 1.0 So, A + C should make the 31st digit to 1, which is surely outside the precision level ... + b; float x = a + b; printf("x = %0.25f\n",x); x = x + c; printf("x = %0.25f\n",x); }
20
bcz in the question itsself given i.e conditional instructions evaluate in the 3rd stage , so there will be 2 stall cycle.